Which is an exponential decay function? f(x) =3/4 (7/4)x f(x) =2/3 (4/5)–x f(x) =3/2 (8/7)–x f(x) =1/3 (9/2)x

first correct answer gets !

f(x) =3/2 (8/7)^(–x) is the only expo. decay function here.

Step-by-step explanation:

First, let's write each of these functions on its own line for greater clarity:

f(x) =3/4 (7/4)x

f(x) =2/3 (4/5)–x

f(x) =3/2 (8/7)–x

f(x) =1/3 (9/2)x

Next, use " ^ " to indicate exponentiation:

f(x) =3/4 (7/4)^x

f(x) =2/3 (4/5)^(–x)

f(x) =3/2 (8/7)^(–x)

f(x) =1/3 (9/2)^x

An exponential growth function looks like f(x) = a(base)^x, where the base is greater than one.

Therefore, we eliminate f(x) =3/4 (7/4)^x; the base (7/4) is greater than 1.

The next one, f(x) =2/3 (4/5)^(–x), is equivalent to f(x) =2/3 (5/4)^x.  This is a growth function because the base is  5/4 (greater than 1).

The next one, f(x) =3/2 (8/7)^(–x), is equivalent to f(x) =3/2 (7/8)^(x).  Here the base is between 0 and 1, so this is the exponential decay function.

The last one has a base greater than 1, so is an expo. growth function.

translate k to n and reflect across the line containing jk.